package Backtracking;

import java.util.Scanner;
/*
 * Problem Statement :
 * Find the number of ways that a given integer, N , can be expressed as the sum of the Xth powers of unique, natural numbers.
 * For example, if N=100 and X=3, we have to find all combinations of unique cubes adding up to 100. The only solution is 1^3+2^3+3^3+4^3.
 * Therefore output will be 1.
*/
public class PowerSum {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		System.out.println("Enter the number and the power");
		int N = sc.nextInt();
		int X = sc.nextInt();
		PowerSum ps = new PowerSum();
		int count = ps.powSum(N,X);
		//printing the answer.
		System.out.println("Number of combinations of different natural number's raised to "+X+" having sum "+N+" are : ");
		System.out.println(count);
		sc.close();
	}
	private int count = 0,sum=0;
    public int powSum(int N, int X) {
        Sum(N,X,1);
        return count;
    }
    //here i is the natural number which will be raised by X and added in sum.
    public void Sum(int N, int X,int i) {
    	//if sum is equal to N that is one of our answer and count is increased.
        if(sum == N) {
        	count++;
        	return;
        }
        //we will be adding next natural number raised to X only if on adding it in sum the result is less than N.
        else if(sum+power(i,X)<=N) {
        	sum+=power(i,X);
            Sum(N,X,i+1);
            //backtracking and removing the number added last since no possible combination is there with it.
            sum-=power(i,X);
        }
        if(power(i,X)<N) {
        	//calling the sum function with next natural number after backtracking if when it is raised to X is still less than X.
        	Sum(N,X,i+1);
        }
    }
    //creating a separate power function so that it can be used again and again when required. 
    private int power(int a , int b ){
        return (int)Math.pow(a,b);
    }
}
